Question

Sinx+sin^2x+cos^3x=0

Answer 1

$\sin x+\sin^2x+\cos^3x=0\\ \sin x+\sin^2x+\cos^2x\cos x=0\\ \sin x(1+\sin x)+(1-\sin x)(1+\sin x)\cos x=0\\ (1+\sin x)(\sin x+\cos x-\sin x\cos x)=0$

Произведение равно нулю, если хотя бы один из множителей равен нулю

$1+\sin x=0\\ \sin x=-1\\ x=-\frac{\pi}{2} +2\pi k,k \in \mathbb{Z}$

$\sin x+\cos x-\sin x\cos x=0$

Пусть $\sin x+\cos x=t~~(|t|\leq \sqrt{2} )$, тогда возводим в квадрат обе части равенства, получим $1+2\sin x\cos x=t^2~~\Rightarrow~~ \sin x\cos x=\frac{t^2-1}{2}$

$t+\frac{t^2-1}{2} =0\\ t^2+2t-1=0\\ D=b^2-4ac=2^2-4\cdot1\cdot(-1)=4+4=8\\ \\ t_1=\dfrac{-2+2\sqrt{2}}{2} =-1+\sqrt{2} \\ t_2=-1-\sqrt{2} ~~\notin~~|t|\leq \sqrt{2}$

обратная замена

$\sin x+\cos x=-1+\sqrt{2} \\ \sqrt{2} \sin (x+\frac{\pi}{4} )=-1+\sqrt{2} \\ \\ \sin (x+\frac{\pi}{4})=-\frac{1}{\sqrt{2}} +1\\ \\ x=(-1)^k\cdot\arcsin(1-\frac{1}{\sqrt{2}})-\frac{\pi}{4}+\pi k,k \in \mathbb{Z}$

Answer 2

To solve the equation sin(x) + sin^2(x) + cos^3(x) = 0, we can simplify it and then attempt to find its solutions. Let's break it down step by step:

1. Start with the equation: sin(x) + sin^2(x) + cos^3(x) = 0.

2. Use the trigonometric identity sin^2(x) = 1 - cos^2(x). Replace sin^2(x) with this identity in the equation:

   sin(x) + (1 - cos^2(x)) + cos^3(x) = 0.

3. Rearrange the terms:

   cos^3(x) - cos^2(x) + sin(x) + 1 = 0.

4. Now, we can attempt to solve this equation. Unfortunately, it doesn't have a simple algebraic solution. We'll need to use numerical methods or approximations to find the solutions.

   One common approach is to use numerical techniques such as graphing or iterative methods like Newton's method to approximate the values of x that satisfy the equation.

   If you provide a specific range or context for the solutions you're seeking, I can assist you further in finding approximations or using numerical methods to solve the equation.